Timeline for Reduction of different RG lattices to kG modules
Current License: CC BY-SA 3.0
4 events
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| Apr 4, 2014 at 17:18 | comment | added | Geoff Robinson | Yes, that is the relevant result (or its dual). The reduced lattice has the same head as the PIMdoes, and that head is simple. A module with simple head must be indecomposable. The way to dualise it is to take the lattice affording the irreducible character as a pure submodule of the "lift" of the PIM. Then its reduction has the same socle as the PIM, which is simple. | |
| Apr 4, 2014 at 17:04 | comment | added | daveh | I've just looked at the Thompson paper and I don't see that result. The only general result is Theorem 1, and that seems to show that we choose any subcharacter of a PIM and find a lattice so it appears at the top when we reduce, but nothing about Brauer characters. | |
| Apr 4, 2014 at 16:18 | comment | added | daveh | Thanks, I had looked in Curtis & Reiner, Nagao-Tsushima, Benson but not in Feit! | |
| Apr 4, 2014 at 14:58 | history | answered | Geoff Robinson | CC BY-SA 3.0 |